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Schach [20]
3 years ago
9

PL-1) A spring that hangs vertically is 25 cm long when no weight is attached to its lower end. Steve adds 250 g of mass to the

end of the spring, which stretches to a new length of 37 cm. What is the spring constant, k, in N/m?
Physics
1 answer:
Anuta_ua [19.1K]3 years ago
7 0

Answer:

20.42 N/m

Explanation:

From hook's law,

F = ke ......................... Equation 1

Where F = Force applied to the spring., k = spring constant, e = extension.

Make k the subject of the equation,

k = F/e ................. Equation 2

Note: The force on the spring is equal to the weight of the mass hung on it.

F = W = mg.

k = mg/e................ Equation 3

Given: m = 250 g = 0.25 kg, e = 37-25 = 12 cm = 0.12 m.

Constant: g = 9.8 m/s²

Substitute into equation 3

k = (0.25×9.8)/0.12

k = 20.42 N/m.

Hence the spring constant = 20.42 N/m

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The velocity vector of the planet points toward the center of the  circle is the following is true about a planet orbiting a star in uniform circular  motion.

A. The velocity vector of the planet points toward the center of the  circle.

<u>Explanation:</u>

Motion of the planet around the star is mentioned to be uniform and around a circular path. Objects in uniform circular motion motion has constant angular speed but the velocity of the object will not remain constant. Since the planet is in circular motion the direction of velocity vector at a particular point is tangential to the circular path at that particular point.

Thus at every point, the direction of velocity vector changes and this means the velocity is never constant. The objects in uniform circular motion has centripetal acceleration which means that velocity vector of the planet points toward the center of the  circle.

7 0
3 years ago
Which of the following involve an increase in the entropy of the system under consideration? a. melting of a solid b. evaporatio
Mama L [17]

Explanation:

Since, entropy is the measure of degree of randomness. So, more randomly the molecules of a substance are moving more will be its entropy.

  • For example, when a solid melts then it means heat is absorbed by it due to which its molecules have gained energy. As a result, they collide with each other and hence, entropy will increase.
  • Evaporation of a liquid will also cause the liquid to change its state from liquid to gas. This means molecules will go far away from each other leading to an increase in the entropy.
  • Sublimation is a process of conversion of a solid into gaseous phase without going through liquid phase. So, in this case also entropy will increase due to gain in energy by the molecules of a solid.
  • In freezing, molecules of a substance come closer to each other and acquire less energy. Hence, entropy decreases.
  • Mixing is a process of combining two or more substances physically with each other. This leads to increase in entropy of a substance.
  • In separation molecules are separated from each other leading to a decrease in energy. Hence, entropy will also decrease.
  • Diffusion is a process in which molecules are able to rapidly move from one place to another. Hence, entropy increases when diffusion takes place.

Thus, we can conclude that melting of a solid, evaporation of a liquid, sublimation, mixing and diffusion involve an increase in the entropy of the system under consideration.

8 0
3 years ago
A horizontal 2.00\ m2.00 m long, 5.00\ kg5.00 kg uniform beam that lies along the east-west direction is acted on by two forces.
Sunny_sXe [5.5K]

Answer: 240\ rad/s^2

Explanation:

Given

Length of beam l=2\ m

mass of beam m=5\ kg

Two forces of equal intensity acted in the opposite direction, therefore, they create a torque of magnitude

\tau =F\times l=200\times 2=400\ N.m

Also, the beam starts rotating about its center

So, the moment of inertia of the beam is

I=\dfrac{ml^2}{12}=\dfrac{5\times 2^2}{12}\\\\I=\dfrac{5}{3}\ kg.m^2

Torque is the product of moment of inertia and angular acceleration

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7 0
3 years ago
F = 2.0*10^20 N,
natka813 [3]

F=\dfrac{Gm_1m_2}{r^2}

With the given values of F,G,m_1,r, we have

2.0\times10^{20}\,\mathrm N=\dfrac{\left(6.67\times10^{-11}\,\frac{\mathrm{Nm}^2}{\mathrm{kg}^2}\right)\left(5.98\times10^{24}\,\mathrm{kg}\right)m_2}{\left(3.8\times10^8\,\mathrm m\right)^2}

Try dealing with the powers of 10 first: On the right, we have

\dfrac{10^{-11}\times10^{24}}{(10^8)^2}=\dfrac{10^{24-11}}{10^{16}}=10^{-3}

Meanwhile, the other values on the right reduce to

\dfrac{6.67\times5.98}{3.8^2}\approx2.76

Then taking units into account, we end up with the equation

2.0\times10^{20}\,\mathrm N=\left(2.76\times10^{-3}\,\dfrac{\mathrm N}{\mathrm{kg}}\right)m_2

Now we solve for m_2:

m_2=\dfrac{2.0\times10^{20}\,\mathrm N}{2.76\times10^{-3}\,\frac{\mathrm N}{\mathrm{kg}}}\approx0.725\times10^{20-(-3)}\,\mathrm{kg}

m_2=7.25\times10^{22}\,\mathrm{kg}

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m_2=7.3\times10^{22}\,\mathrm{kg}

5 0
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tresset_1 [31]

Answer:

7657

Explanation:

3 0
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